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An introduction to the general properties of the Fourier series

Introduction

In this module we will discuss the basic properties of the Continuous-Time Fourier Series. We will begin by refreshing your memory of our basic Fourier series equations:

f t n c n ω 0 n t
c n 1 T t 0 T f t ω 0 n t
Let · denote the transformation from f t to the Fourier coefficients f t n n c n · maps complex valued functions to sequences of complex numbers .

Linearity

· is a linear transformation .

If f t c n and g t d n . Then α α α f t α c n and f t g t c n d n

Easy. Just linearity of integral.

f t g t n n t 0 T f t g t ω 0 n t n n 1 T t 0 T f t ω 0 n t 1 T t 0 T g t ω 0 n t n n c n d n c n d n

Shifting

Shifting in time equals a phase shift of Fourier coefficients

f t t 0 ω 0 n t 0 c n if c n c n c n , then ω 0 n t 0 c n ω 0 n t 0 c n c n ω 0 t 0 n c n ω 0 t 0 n

f t t 0 n n 1 T t 0 T f t t 0 ω 0 n t n n 1 T t t 0 T t 0 f t t 0 ω 0 n t t 0 ω 0 n t 0 n n 1 T t t 0 T t 0 f t ~ ω 0 n t ~ ω 0 n t 0 n n ω 0 n t ~ c n

Parseval's relation

t 0 T f t 2 T n c n 2
Parseval's relation tells us that the energy of a signal is equal to the energy of its Fourier transform.
Parseval tells us that the Fourier series maps L 0 T 2 to l 2 .

For f t to have "finite energy," what do the c n do as n ?

c n 2 for f t to have finite energy.

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If n n 0 c n 1 n , is f L 0 T 2 ?

Yes, because c n 2 1 n 2 , which is summable.

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Now, if n n 0 c n 1 n , is f L 0 T 2 ?

No, because c n 2 1 n , which is not summable.

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The rate of decay of the Fourier series determines if f t has finite energy .

Parsevals theorem demonstration

ParsevalsDemo
Interact (when online) with a Mathematica CDF demonstrating Parsevals Theorem. To download, right click and save file as .cdf.

Symmetry properties

Even signals

    Even signals

  • f ( t ) = f ( - t )
  • c n = c - n
  • c n = 1 T 0 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( t ) exp ( - ı ω 0 n t ) d t + 1 T T 2 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( - t ) exp ( - ı ω 0 n t ) d t + 1 T T 2 T f ( - t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T f ( t ) exp ( ı ω 0 n t ) d t + exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T f ( t ) 2 cos ( ω 0 n t ) d t

Odd signals

    Odd signals

  • f ( t ) = -f ( -t )
  • c n = c - n *
  • c n = 1 T 0 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( t ) exp ( - ı ω 0 n t ) d t + 1 T T 2 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( t ) exp ( - ı ω 0 n t ) d t - 1 T T 2 T f ( - t ) exp ( ı ω 0 n t ) d t
  • = - 1 T 0 T f ( t ) exp ( ı ω 0 n t ) d t - exp ( - ı ω 0 n t ) d t
  • = - 1 T 0 T f ( t ) 2 ı sin ( ω 0 n t ) d t

Real signals

    Real signals

  • f ( t ) = f * ( t )
  • c n = c - n *
  • c n = 1 T 0 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( t ) exp ( - ı ω 0 n t ) d t + 1 T T 2 T f ( t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T 2 f ( - t ) exp ( - ı ω 0 n t ) d t + 1 T T 2 T f ( - t ) exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T f ( t ) exp ( ı ω 0 n t ) d t + exp ( - ı ω 0 n t ) d t
  • = 1 T 0 T f ( t ) 2 cos ( ω 0 n t ) d t

Differentiation in fourier domain

f t c n t f t n ω 0 c n

Since

f t n c n ω 0 n t
then
t f t n c n t ω 0 n t n c n ω 0 n ω 0 n t
A differentiator attenuates the low frequencies in f t and accentuates the high frequencies. It removes general trends and accentuates areas of sharpvariation.
A common way to mathematically measure the smoothness of a function f t is to see how many derivatives are finite energy.
This is done by looking at the Fourier coefficients of thesignal, specifically how fast they decay as n .If f t c n and c n has the form 1 n k , then t m f t n ω 0 m c n and has the form n m n k .So for the m th derivative to have finite energy, we need n n m n k 2 thus n m n k decays faster than 1 n which implies that 2 k 2 m 1 or k 2 m 1 2 Thus the decay rate of the Fourier series dictates smoothness.

Fourier differentiation demonstration

FourierDiffDemo
Interact (when online) with a Mathematica CDF demonstrating Differentiation in the Fourier Domain. To download, right click and save file as .cdf.

Integration in the fourier domain

If

f t c n
then
τ t f τ 1 ω 0 n c n
If c 0 0 , this expression doesn't make sense.

Integration accentuates low frequencies and attenuates high frequencies. Integrators bring out the general trends in signals and suppress short term variation (which is noise in many cases). Integrators are much nicer than differentiators.

Fourier integration demonstration

fourierIntDemo
Interact (when online) with a Mathematica CDF demonstrating Integration in the Fourier Domain. To download, right click and save file as .cdf.

Signal multiplication and convolution

Given a signal f t with Fourier coefficients c n and a signal g t with Fourier coefficients d n , we can define a new signal, y t , where y t f t g t . We find that the Fourier Series representation of y t , e n , is such that e n k c k d n - k . This is to say that signal multiplication in the time domainis equivalent to signal convolution in the frequency domain, and vice-versa: signal multiplication in the frequency domain is equivalent to signal convolution in the time domain.The proof of this is as follows

e n 1 T t 0 T f t g t ω 0 n t 1 T t 0 T k c k ω 0 k t g t ω 0 n t k c k 1 T t 0 T g t ω 0 n k t k c k d n - k
for more details, see the section on Signal convolution and the CTFS

Conclusion

Like other Fourier transforms, the CTFS has many useful properties, including linearity, equal energy in the time and frequency domains, and analogs for shifting, differentation, and integration.

Properties of the ctfs
Property Signal CTFS
Linearity a x ( t ) + b y ( t ) a X ( f ) + b Y ( f )
Time Shifting x ( t - τ ) X ( f ) e - j 2 π f τ / T
Time Modulation x ( t ) e j 2 π f τ / T X ( f - k )
Multiplication x ( t ) y ( t ) X ( f ) * Y ( f )
Continuous Convolution x ( t ) * y ( t ) X ( f ) Y ( f )

Questions & Answers

show that the set of all natural number form semi group under the composition of addition
Nikhil Reply
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Lukman Reply
_3_2_1
felecia
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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
SABAL Reply
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
Pawel
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Ifeanyi
on number 2 question How did you got 2x +2
Ifeanyi
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Pawel
x*x=2
felecia
2+2x=
felecia
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
mariel Reply
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
Pawel
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Mark
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Abdullahi Reply
find the value of 2x=32
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corri
X=16
Michael
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Only Reply
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4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
Sidiki Reply
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Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
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Need help solving this problem (2/7)^-2
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x+2y-z=7
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-1
Shedrak
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Source:  OpenStax, Signals and systems. OpenStax CNX. Aug 14, 2014 Download for free at http://legacy.cnx.org/content/col10064/1.15
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